Question

Determine the zeros of the quadratic function y = x^2 + 15x + 26. (When listing the smaller zero, remember that negative values are always smaller than positive values.)

Answer 1

To determine the zeros of the quadratic function \( y = x^2 + 15x + 26 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1 \), \( b = 15 \), and \( c = 26 \).

First, we need to calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = 15^2 = 225
\]

\[
4ac = 4 \times 1 \times 26 = 104
\]

Now, compute the discriminant:

\[
b^2 - 4ac = 225 - 104 = 121
\]

Since the discriminant is positive, we will have two distinct real zeros.

Now, substitute back into the quadratic formula:

\[
x = \frac{-15 \pm \sqrt{121}}{2 \cdot 1}
\]

Simplifying this further:

\[
\sqrt{121} = 11
\]

Now we have:

\[
x = \frac{-15 \pm 11}{2}
\]

Calculating the two values:

1. For \( x = \frac{-15 + 11}{2} \):

\[
x = \frac{-4}{2} = -2
\]

2. For \( x = \frac{-15 - 11}{2} \):

\[
x = \frac{-26}{2} = -13
\]

Thus, the two zeros of the quadratic function \( y = x^2 + 15x + 26 \) are:

\[
x = -13 \quad \text{and} \quad x = -2
\]

In summary, the smaller zero is \( -13 \) and the larger zero is \( -2 \).

Therefore, the final answer is:

\[
\text{Zeros: } -13, -2
\]